Question: A particle moves along the $x$ -axis. The function $v(t)$ gives the particle's velocity at any time $t\geq 0$ : $v(t)=t^3-3t^2-8t+3$ What is the particle's velocity $v(t)$ at $t=4$ ? $v(4)=$
Solution: In the first part, we have a function for the particle's velocity, and we need to find the particle's velocity. So all we have to do is evaluate the velocity function. In the second part, we need to find the particle's acceleration. Since acceleration is the rate of change of velocity, we need to find the derivative of $v(t)$. In other words, if $a(t)$ gives the particle's acceleration at any time $t\geq 0$, then $a(t)=v'(t)$. First, let's evaluate $v(4)$ : $\begin{aligned} v({4})&=({4})^3-3({4})^2-8({4})+3 \\\\ &=-13 \end{aligned}$ Notice that the velocity is negative. This means the particle is moving to the left. Now, let's differentiate $v(t)$ to find $a(t)$ : $\begin{aligned} a(t)&=v'(t) \\\\ &=\dfrac{d}{dt}[t^3-3t^2-8t+3] \\\\ &=3t^2-6t-8 \end{aligned}$ To find the particle's acceleration at $t=4$, we need to evaluate $a(4)$. $\begin{aligned} a({4})&=3({4})^2-6({4})-8 \\\\ &=16 \end{aligned}$ Since the particle is accelerating toward the right while moving toward the left, we know the particle is slowing down. The particle's velocity at $t=4$ is $-13$. The particle's acceleration at $t=4$ is $16$. At $t=4$, the particle is slowing down.